Question: A curve in the plane is defined parametrically by the equations $x=t^2-1$ and $y=2e^t$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2e^t(t^2-1-2t)}{(t^2-1)^2}$ (Choice B) B $\dfrac{e^t}{t}$ (Choice C) C $\dfrac{2e^t}{t}$ (Choice D) D $2e^t$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=t^2-1$ and $y=2e^t$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}(2e^t)}{\dfrac{d}{dt}(t^2-1)} \\\\ &=\dfrac{2e^t}{2t} \\\\ &=\dfrac{e^t}{t} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{e^t}{t}$.